The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of aluminium = 
= specific heat of water = 
= mass of aluminum = 0.500 kg = 500 g
= mass of water = 0.250 kg = 250 g
= final temperature of mixture = ?
= initial temperature of aluminum = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 
Answer:
60 N
Explanation:
This is just Newton's Second Law
F = m*a
F = ?
m = 12 kg
a = 5 m/^2
F = 5*12 = 60 Newtons
The Bohr's proposal for the angular momentum of an electron in Bohr's model of the hydrogen atom is:
L=(n*h)/(2π), where n is the number of the energy level and h is the Planck's constant. This equation shows us the quantization of angular momentum of the electron. So the correct answer is the second one: Planck's constant.
The meter out circuit is the flow control circuit design that can most effectively control an overrunning load.
The meter-out circuit can be very accurate, but are not efficient. The meter-out circuit can control overrunning as well as opposing loads while the other one method must be used with opposing loads only. The choice of flown control valve method and the location of the flow control in the circuit are dependent on the type of application being controlled.
<h3>What is a Circuit ?</h3>
In electronics, a circuit is a complete circular conduit through which electricity flows. A simple circuit consists of conductors, a load, and a current source. The term "circuit" broadly refers to any continuous path via which electricity, data, or a signal might flow.
- The directional valve shifts, causing the actuator to move faster than pump flow can fill it due to an overrunning load. Oil is leaking from one side, whereas there is none on the other.
Hence, flow control circuit design that can best control an overrunning load is the opposing circuit
Learn more about Circuit here:
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